Largest Product
the n value that maximises $(\frac{k}{n})^n$ is $k/e$ though
interestingly, this cannot be obtained by calculus (says Mike)
this is clearly beyond the scope of most stage three students (non
integer powers)though they may well graph the function
$(\frac{k}{n})^n$ for some k, or on autograph with varying k
these are the student solutions from the original publication of
the problem :
When $10$ is divided into $2$ parts, the largest product is $5
\times5 = 25$.
If it is divided into $3$ parts, the largest product is $3
{1\over3} \times3 {1\over3} \times3 {1\over3}$.
If it is divided into $4$ parts, the largest product is $2
{1\over2}\times 2 {1\over2} \times2 {1\over2} \times2
{1\over2}$.
If it is divided into $5$ parts, the largest product is $2\times 2
\times2 \times2 \times2$.
It soon becomes apparent that when $10$ is split into $n$ parts
(where n is a whole number), the largest product is given by
$\left({10\over n}\right)^n$.
So the problem is in fact to find the largest value of
$\left({10\over{n}}\right)^n$.
The largest value of $\left({10\over{n}}\right)^n$ is when
$n=4$:
$\left({10\over4}\right)^4 = 2.5^4 = 39.0625$
This is the largest product.
Here is an explanation to why this works: if there are two numbers
added to make $10$ they can be thought of as making a rectangle and
by plotting length against height for different rectangles on a
graph it is possible to see that the maximum area occurs (the
product) when the length equals the breadth.
The same type of argument can be made for three numbers forming a
cuboid. The greatest volume will occur when the three dimensions of
the cuboid are equal (a cube).
So we can definitely show that this is the case for sums made from
two and three numbers.
Correct solutions to this were sent in by:
| Hannah |
Archbishop Sancroft High
School |
| Stephen |
Aylsham Middle School |
|
|
|